YES

We show the termination of the TRS R:

  -(|0|(),y) -> |0|()
  -(x,|0|()) -> x
  -(x,s(y)) -> if(greater(x,s(y)),s(-(x,p(s(y)))),|0|())
  p(|0|()) -> |0|()
  p(s(x)) -> x

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: -#(x,s(y)) -> -#(x,p(s(y)))
p2: -#(x,s(y)) -> p#(s(y))

and R consists of:

r1: -(|0|(),y) -> |0|()
r2: -(x,|0|()) -> x
r3: -(x,s(y)) -> if(greater(x,s(y)),s(-(x,p(s(y)))),|0|())
r4: p(|0|()) -> |0|()
r5: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: -#(x,s(y)) -> -#(x,p(s(y)))

and R consists of:

r1: -(|0|(),y) -> |0|()
r2: -(x,|0|()) -> x
r3: -(x,s(y)) -> if(greater(x,s(y)),s(-(x,p(s(y)))),|0|())
r4: p(|0|()) -> |0|()
r5: p(s(x)) -> x

The set of usable rules consists of

  r5

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        -#_A(x1,x2) = max{x1 + 6, x2 + 2}
        s_A(x1) = max{8, x1 + 5}
        p_A(x1) = max{2, x1 - 4}
  
    precedence:
    
      -# = s = p
    
    partial status:
  
      pi(-#) = [2]
      pi(s) = []
      pi(p) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.